Question

Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks.
Construct a 95% confience interval for the percentage of executives that preferred trucks. ( I know for sure that the answer is +or- 6.82%, but I need to know how to get this.)

Answer 1

Answer:

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

Step-by-step explanation:

We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.

We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.26.

 

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26*0.74}{160}}\\\\\\ \sigma_p=\sqrt{0.0012}=0.0347$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0347=0.068$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.26-0.068=0.192\\\\UL=p+z \cdot \sigma_p = 0.26+0.068=0.328$

The 95% confidence interval for the population proportion is (0.192, 0.328).

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.